Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $a = \dfrac{y^2 - 25}{y^2 + 10y} \times \dfrac{-8y - 80}{-8y - 40} $
Solution: First factor the quadratic. $a = \dfrac{(y + 5)(y - 5)}{y^2 + 10y} \times \dfrac{-8y - 80}{-8y - 40} $ Then factor out any other terms. $a = \dfrac{(y + 5)(y - 5)}{y(y + 10)} \times \dfrac{-8(y + 10)}{-8(y + 5)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ (y + 5)(y - 5) \times -8(y + 10) } { y(y + 10) \times -8(y + 5) } $ $a = \dfrac{ -8(y + 5)(y - 5)(y + 10)}{ -8y(y + 10)(y + 5)} $ Notice that $(y + 10)$ and $(y + 5)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ -8\cancel{(y + 5)}(y - 5)(y + 10)}{ -8y(y + 10)\cancel{(y + 5)}} $ We are dividing by $y + 5$ , so $y + 5 \neq 0$ Therefore, $y \neq -5$ $a = \dfrac{ -8\cancel{(y + 5)}(y - 5)\cancel{(y + 10)}}{ -8y\cancel{(y + 10)}\cancel{(y + 5)}} $ We are dividing by $y + 10$ , so $y + 10 \neq 0$ Therefore, $y \neq -10$ $a = \dfrac{-8(y - 5)}{-8y} $ $a = \dfrac{y - 5}{y} ; \space y \neq -5 ; \space y \neq -10 $